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Question Number 20786 by Tinkutara last updated on 02/Sep/17

$$\mathrm{Two}\mathrm{particles}A\mathrm{and}B\mathrm{start}\mathrm{from}\mathrm{the}$$$$\mathrm{same}\mathrm{position}\mathrm{along}\mathrm{the}\mathrm{circular}\mathrm{path}\mathrm{of}$$$$\mathrm{radius}0.5\mathrm{m}\mathrm{with}\mathrm{a}\mathrm{speed}{v}_A=1{\mathrm{ms}}^{-1}$$$$\mathrm{and}{v}_B=1.2{\mathrm{ms}}^{-1}\mathrm{in}\mathrm{opposite}\mathrm{direction}.$$$$\mathrm{Determine}\mathrm{the}\mathrm{time}\mathrm{before}\mathrm{they}\mathrm{collide}.$$

Answered by dioph last updated on 03/Sep/17

$${\theta }_A={\omega }_{A}t=\frac{v_{A}t}{r}=\frac{1\times t}{0.5}=2t$$$${\theta }_B=2\pi -{\omega }_{B}t=2\pi -\frac{v_{B}t}{r}=2\pi -\frac{1.2\times t}{0.5}=2\pi -2.4t$$$${\theta }_A={\theta }_B\iff 2t=2\pi -2.4t\iff$$$$\iff 4.4t=2\pi \iff t=\frac{2\pi }{4.4}s$$

Commented by Tinkutara last updated on 03/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Answered by mrW1 last updated on 03/Sep/17

$$\mathrm{t}=\frac{2\pi \mathrm{r}}{{\mathrm{v}}_{\mathrm{A}}+{\mathrm{v}}_{\mathrm{B}}}=\frac{2\pi \times 0.5}{1+1.2}=\frac{\pi }{2.2}=1.43\mathrm{s}$$

Commented by Tinkutara last updated on 03/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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